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Analytical Geometry

Grade 10 CAPS: the distance, gradient and midpoint formulas, and the conditions for parallel and perpendicular lines — the toolkit every later grade builds on.

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Grade 10 CAPS Mathematics

Analytical Geometry

Four formulas, one Cartesian plane: distance, gradient, midpoint, and the parallel/perpendicular test. Every worked example below moves from a direct substitution up to a genuine application, so work through them in order.

Scope check:Official CAPS requirement

Your 4-hour learning route

Four formulas, each taught the same way: concept, then examples from simple to hard.

55
min
1. Distance formula

How far apart are two points?

50
min
2. Gradient formula

How steep is the line between them?

50
min
3. Parallel & perpendicular

Comparing two gradients.

45
min
4. Midpoint formula

Finding the exact centre.

40
min
5. Combining the toolkit

Classifying shapes on the Cartesian plane.

Study rule
Every formula here takes two points, \((x_1,y_1)\) and \((x_2,y_2)\), and returns one number (or one point). Get comfortable labelling which point is 1 and which is 2 — it never changes the distance or midpoint answer, but it DOES flip the sign of the gradient if you're inconsistent mid-calculation.

What CAPS actually asks in Grade 10

Four formulas, applied to any two points on the Cartesian plane.

  • 1

    Represent geometric figures in a Cartesian co-ordinate system, and derive and apply, for any two points \((x_1,y_1)\) and \((x_2,y_2)\), a formula for calculating the distance between the two points.

  • 2

    The gradient of the line segment joining the points.

  • 3

    Conditions for parallel and perpendicular lines.

  • 4

    The co-ordinates of the midpoint of the line segment joining the points.

Not yet
Finding the actual equation of a line, and the inclination (angle) of a line, are Grade 11 content. Grade 10 stops at these four formulas.

The distance formula

Built directly from a right-angled triangle and the Theorem of Pythagoras.

Distance between two points
\[d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

Draw the horizontal leg (\(x_2-x_1\)) and the vertical leg (\(y_2-y_1\)) of a right-angled triangle between the two points. The distance IS the hypotenuse — this formula is nothing more than Pythagoras' theorem.

Worked example: the distance formula

Four levels: substitute directly, handle negatives, work backward, then apply it.

Direct substitution Level 1

Calculate the distance between \(A(1,2)\) and \(B(4,6)\).

Show the working
\(d=\sqrt{(4-1)^2+(6-2)^2}=\sqrt{9+16}=\sqrt{25}=\boxed{5}\)
With negative coordinates Level 2

Calculate the distance between \(A(-3,2)\) and \(B(5,-4)\).

Show the working
\(d=\sqrt{(5-(-3))^2+(-4-2)^2}=\sqrt{8^2+(-6)^2}=\sqrt{64+36}=\sqrt{100}=\boxed{10}\)
Working backward for an unknown coordinate Level 3

\(A(2,1)\) and \(B(x,7)\) are 10 units apart. Calculate the possible value(s) of \(x\).

Show the working
  1. 1\(10=\sqrt{(x-2)^2+(7-1)^2}\)
  2. 2\(100=(x-2)^2+36\Rightarrow(x-2)^2=64\)
  3. 3\(x-2=\pm8\Rightarrow\boxed{x=10\text{ or }x=-6}\)
Application: is the triangle isosceles? Level 4

\(P(0,0)\), \(Q(4,3)\) and \(R(8,0)\) are the vertices of a triangle. Show that \(\triangle PQR\) is isosceles.

Show the working
  1. 1\(PQ=\sqrt{(4-0)^2+(3-0)^2}=\sqrt{16+9}=5\)
  2. 2\(QR=\sqrt{(8-4)^2+(0-3)^2}=\sqrt{16+9}=5\)
  3. 3\(\boxed{PQ=QR=5}\), so \(\triangle PQR\) is isosceles — you never needed \(PR\) at all.
Quick check: distance formula

Calculate the distance between \(A(0,0)\) and \(B(3,4)\).

To find an unknown coordinate from a given distance, after substituting you should:

The gradient formula

Rise over run — how steep the line segment is.

Gradient of a line segment
\[m=\dfrac{y_2-y_1}{x_2-x_1}\]

A positive gradient rises left to right; a negative gradient falls. A horizontal line has \(m=0\); a vertical line has an undefined gradient (division by zero).

Worked example: the gradient formula

Direct substitution, then working backward, then proving three points lie on one line.

Direct substitution, positive gradient Level 1

Calculate the gradient of the line through \(A(1,1)\) and \(B(4,7)\).

Show the working
\(m=\dfrac{7-1}{4-1}=\dfrac{6}{3}=\boxed{2}\)
Direct substitution, negative gradient Level 1

Calculate the gradient of the line through \(A(-2,5)\) and \(B(3,-5)\).

Show the working
\(m=\dfrac{-5-5}{3-(-2)}=\dfrac{-10}{5}=\boxed{-2}\)
Working backward for an unknown coordinate Level 3

The line through \(A(2,3)\) and \(B(6,y)\) has a gradient of 2,5. Calculate \(y\).

Show the working
  1. 1\(2{,}5=\dfrac{y-3}{6-2}\Rightarrow2{,}5\times4=y-3\)
  2. 2\(10=y-3\Rightarrow\boxed{y=13}\)
Application: are three points collinear? Level 4

Show that \(A(1,2)\), \(B(3,6)\) and \(C(5,10)\) lie on the same straight line.

Show the working
  1. 1\(m_{AB}=\dfrac{6-2}{3-1}=\dfrac{4}{2}=2\)
  2. 2\(m_{BC}=\dfrac{10-6}{5-3}=\dfrac{4}{2}=2\)
  3. 3\(\boxed{m_{AB}=m_{BC}=2}\), and both pass through the shared point \(B\), so \(A\), \(B\) and \(C\) are collinear.
Why this works
Two line segments that share a point AND have the same gradient must be the same line — there is only one line through a point with a given steepness.
Quick check: gradient formula

Calculate the gradient of the line through \(A(2,5)\) and \(B(4,9)\).

To prove three points A, B and C are collinear, you should show that:

Parallel and perpendicular lines

Two gradients, compared two different ways.

Parallel lines
\[m_1=m_2\]

Parallel lines never meet, because they rise (or fall) at exactly the same rate. Equal gradients — nothing more to check.

Perpendicular lines
\[m_1\times m_2=-1\]

Perpendicular lines meet at a right angle. Their gradients are "negative reciprocals" of each other — flip the fraction and change the sign.

Worked example: parallel and perpendicular lines

Check if two lines are parallel, find a perpendicular gradient, then prove a right angle from four points.

Are two lines parallel? Level 1

Line 1 has a gradient of \(\dfrac{3}{4}\). Line 2 passes through \((0,1)\) and \((4,4)\). Are the lines parallel?

Show the working
\(m_2=\dfrac{4-1}{4-0}=\dfrac{3}{4}\). Since \(m_1=m_2=\dfrac{3}{4}\), \(\boxed{\text{the lines are parallel}}\).
Finding a perpendicular gradient Level 2

A line has a gradient of \(\dfrac{2}{3}\). Calculate the gradient of a line perpendicular to it.

Show the working
\(m_1\times m_2=-1\Rightarrow m_2=\dfrac{-1}{m_1}=\dfrac{-1}{2/3}=\boxed{-\dfrac{3}{2}}\)
Application: proving a right angle Level 4

\(A(0,0)\), \(B(2,4)\), \(C(4,0)\) and \(D(2,1)\) are four points. Show that \(AB\perp CD\).

Show the working
  1. 1\(m_{AB}=\dfrac{4-0}{2-0}=2\)
  2. 2\(m_{CD}=\dfrac{1-0}{2-4}=-0{,}5\)
  3. 3\(m_{AB}\times m_{CD}=2\times(-0{,}5)=\boxed{-1}\), so \(AB\perp CD\).
Common mistake
"Negative reciprocal" means BOTH flip the fraction AND change the sign. A gradient of 2 (i.e. 2/1) becomes −1/2, not just −2 and not just 1/2.
Quick check: parallel & perpendicular

A line has a gradient of 4. What is the gradient of a line perpendicular to it?

Two lines have gradients that multiply to give exactly −1. What can you conclude?

The midpoint formula

The exact centre of a line segment — just average the two coordinates.

Midpoint of a line segment
\[M=\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)\]

Average the two \(x\)-values for the midpoint's \(x\); average the two \(y\)-values for the midpoint's \(y\). That's the whole formula.

Worked example: the midpoint formula

Direct substitution, then working backward to find a missing endpoint.

Direct substitution Level 1

Calculate the midpoint of the line segment joining \(A(2,4)\) and \(B(8,10)\).

Show the working
\(M=\left(\dfrac{2+8}{2},\dfrac{4+10}{2}\right)=\boxed{(5,7)}\)
With negative coordinates Level 2

Calculate the midpoint of the line segment joining \(A(-3,5)\) and \(B(7,-9)\).

Show the working
\(M=\left(\dfrac{-3+7}{2},\dfrac{5+(-9)}{2}\right)=\boxed{(2,-2)}\)
Working backward for a missing endpoint Level 3

\(M(5,7)\) is the midpoint of \(A(1,2)\) and \(B\). Calculate the coordinates of \(B\).

Show the working
  1. 1\(5=\dfrac{1+x_B}{2}\Rightarrow x_B=10-1=9\)
  2. 2\(7=\dfrac{2+y_B}{2}\Rightarrow y_B=14-2=12\)
  3. 3\(\boxed{B(9,12)}\)
Application: do the diagonals bisect each other? Level 4

\(PQRS\) has vertices \(P(0,0)\), \(Q(4,2)\), \(R(6,6)\), \(S(2,4)\). Show that the diagonals \(PR\) and \(QS\) bisect each other.

Show the working
  1. 1Midpoint of \(PR\): \(\left(\dfrac{0+6}{2},\dfrac{0+6}{2}\right)=(3,3)\)
  2. 2Midpoint of \(QS\): \(\left(\dfrac{4+2}{2},\dfrac{2+4}{2}\right)=(3,3)\)
  3. 3\(\boxed{\text{Both diagonals share the same midpoint }(3,3)}\), so they bisect each other.
Why this matters
Diagonals that bisect each other is the defining property of a parallelogram — this exact PQRS shape returns in the next section.
Quick check: midpoint formula

Calculate the midpoint of \(A(0,0)\) and \(B(6,10)\).

If diagonals of a quadrilateral share the same midpoint, what does that prove?

Combining the toolkit

Real exam questions rarely ask for just one formula — they ask you to classify a shape.

A four-step checklist for any shape question
  1. Read what's being asked: a specific side length? A type of quadrilateral? A right angle?
  2. Decide which formula answers that question — distance for lengths, gradient for angles/parallel sides, midpoint for bisecting diagonals.
  3. Calculate only what you need — don't compute every possible distance and gradient if the question only needs two.
  4. State your conclusion in words, referencing the numbers you calculated.

Worked example: classifying a quadrilateral

Using distance AND gradient together to prove a shape is a rhombus.

Prove PQRS is a rhombus Level 4

\(P(0,0)\), \(Q(4,2)\), \(R(6,6)\) and \(S(2,4)\) are the vertices of quadrilateral \(PQRS\). Prove that \(PQRS\) is a rhombus.

Show the complete proof
  1. 1\(PQ=\sqrt{4^2+2^2}=\sqrt{20}\)
  2. 2\(QR=\sqrt{2^2+4^2}=\sqrt{20}\)
  3. 3\(RS=\sqrt{4^2+2^2}=\sqrt{20}\)
  4. 4\(SP=\sqrt{2^2+4^2}=\sqrt{20}\)
  5. 5\(\boxed{PQ=QR=RS=SP=\sqrt{20}}\) — all four sides are equal, so \(PQRS\) is at least a rhombus (a special parallelogram).
Bonus check
A genuine rhombus also has perpendicular diagonals: \(m_{PR}=1\) and \(m_{QS}=-1\), and \(1\times(-1)=-1\) confirms it — a second, independent way to reach the same conclusion.

Worked example: classifying a triangle

Using gradient to prove a right angle, then naming the triangle fully.

Is the triangle right-angled? Level 4

\(A(0,0)\), \(B(2,4)\) and \(C(6,2)\) are the vertices of \(\triangle ABC\). Show that \(\triangle ABC\) is right-angled at \(B\).

Show the working
  1. 1\(m_{AB}=\dfrac{4-0}{2-0}=2\)
  2. 2\(m_{BC}=\dfrac{2-4}{6-2}=\dfrac{-2}{4}=-0{,}5\)
  3. 3\(m_{AB}\times m_{BC}=2\times(-0{,}5)=\boxed{-1}\), so \(AB\perp BC\) and \(\triangle ABC\) is right-angled at \(B\).
Read the question carefully
"Right-angled at B" specifically means the two sides MEETING at B (that's AB and BC) must be perpendicular — not AC, which doesn't touch B at all.
Quick check: combining the toolkit

To prove a quadrilateral is a rhombus using the distance formula, you should show that:

To show a triangle is right-angled at a specific vertex B, which two gradients must you multiply?

Mistake clinic: repair the exact error

Small slips that break an otherwise correct answer.

Distance

Order doesn't matter, but consistency does

\((x_2-x_1)\) or \((x_1-x_2)\) give the same squared result — but never mix which point is 1 in the x-part and which is 1 in the y-part.

Gradient

Order DOES matter here

Unlike distance, swapping which point is "1" and which is "2" in the gradient formula flips the sign if you're inconsistent between numerator and denominator.

Perpendicular

Flip AND change sign

The negative reciprocal of \(m\) is \(-\dfrac{1}{m}\) — both steps are required, not just one.

Midpoint

Average, don't add

The midpoint formula divides the SUM of the coordinates by 2 — forgetting the division doubles both coordinates.

Shape proofs

Calculate only what's asked

"Right-angled at B" needs the two sides meeting at B, not all three sides of the triangle.

Your Paper 2 checklist

Use this before submitting an exam answer.

Avoid this

Mixing up which point is 1 and which is 2 mid-calculation.

Forgetting to flip AND negate for a perpendicular gradient.

Calculating more distances/gradients than the question actually needs.

Do this

Label your two points clearly before substituting into any formula.

State your final conclusion in words, not just a number.

Sketch the points roughly on a grid first — a sketch catches an obviously wrong answer immediately.

Calculate the distance between two points, including working backward for an unknown coordinate.
Calculate the gradient of a line segment, including proving collinearity.
Test whether two lines are parallel or perpendicular, and find a missing gradient.
Calculate the midpoint of a line segment, including finding a missing endpoint.
Combine these formulas to classify a triangle or quadrilateral.
More explanation and exercises:Siyavula Grade 10 Analytical Geometry
Summary complete

You now have the toolkit.

Use the Mastery Bank to build fluency. Then take the test without notes and use the result to choose the exact slide to revisit.

Next step 1

Mastery Bank

17 questions by skill, from a direct distance calculation through to classifying a quadrilateral. Answers reveal only after your attempt.

Next step 2

Quick Test

A short exam-style self-check. Mark it, then revisit the exact slide that matches a missed skill.

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Learn It in Short Videos

Four free, independent videos — not made by Equation Station SA.

Distance Formula

Where the formula comes from, and how to use it.

Khan Academy · Distance formula

Gradient (Slope) Formula

Calculating slope between two points, step by step.

Khan Academy · Slope from two ordered pairs

Parallel & Perpendicular Gradients

Recognising and calculating gradients for parallel and perpendicular lines.

Khan Academy · Equations of parallel and perpendicular lines

Midpoint Formula

Finding the midpoint between two coordinates.

Khan Academy · Midpoint formula

Practise in the right order

The core teaching is above. These are the next steps, not a replacement for it.

01
Built-in practice
Analytical Geometry Mastery Bank

17 questions by skill, with concise reveal answers and methods.

Start after the slides
Open Mastery Bank
02
Built-in check
Test Your Knowledge

Use the short exam-style self-check when you want a fast confidence check.

Then target one weak skill
Take the Quick Test
CAPS
Free textbook chapter
Siyavula: Grade 10 Analytical Geometry

Use its own worked examples for extra explanation and exercises.

Free • CAPS aligned
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DBE
Official free books
DBE Grade 10 Textbooks

Official state-owned learner books and teacher support for Grade 10 Mathematics.

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Related Topics

This topic continues into Grade 11 and Grade 12.

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Analytical Geometry (Grade 11)

Equation of a line, inclination

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Analytical Geometry (Grade 12)

Circles and tangents

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Functions & Graphs (Grade 10)

The Cartesian plane and graphing

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Student Questions

Frequently Asked Questions

Short answers for the checks learners make while preparing for the Grade 10 CAPS exam.

What does CAPS require for Grade 10 Analytical Geometry?

Represent geometric figures in a Cartesian co-ordinate system, and derive and apply, for any two points, a formula for calculating the distance between the two points, the gradient of the line segment joining the points, conditions for parallel and perpendicular lines, and the co-ordinates of the midpoint of the line segment joining the points.

Is the equation of a line part of Grade 10?

No. Finding the equation of a line and the inclination of a line are Grade 11 content. Grade 10 covers only distance, gradient, midpoint, and parallel/perpendicular conditions.

How do you prove three points are collinear?

Show that the gradient between the first pair of points equals the gradient between the second pair (using a shared point). Equal gradients through a common point mean all three points lie on the same straight line.

Where should I practise next?

Finish the interactive slides, open the Mastery Bank, then take the short self-test without notes.